Chemistry 121 Oregon State University

Chemistry 121 Oregon State University Worksheet 5 Notes Dr. Richard L Nafshun

A. A student obtains 4.000 moles of carbon atoms. How many carbon atoms are present? What is the mass of this sample?

 23  6.022x10 atoms 24 4.000 moles C atoms   = 2.408 x 10 C atoms  1 mole 

12.011 g  4.000 moles C atoms   = 48.04 g carbon  1 mole 

1. Determine the molar masses (AKA atomic masses) of C, N, O, F, Ne, and U.

C = 12.01 g/mol.

N = 14.01 g/mol.

O = 16.00 g/mol.

F = 19.00 g/mol.

Ne = 20.18 g/mol.

U = 238.03 g/mol.

2. Determine the molar masses of methane, the O2 molecule, carbon dioxide, and water.

Methane (CH4):

1 x 12.01 g/mol + 4 x 1.01 g/mol ______

16.05 g/mol

2 x 16.00 g/mol = 32.00 g/mol

1 x 12.01 g/mol + 2 x 16.00 g/mol ______

44.01 g/mol

2 x 1.01 g/mol + 1 x 16.00 g/mol ______

18.02 g/mol

3. Determine the molar mass of elemental aluminum ("just plain aluminum metal"), HCl, aluminum chloride, and the H2 molecule.

Al is 26.98 g/mol.

HCl is 36.46 g/mol.

AlCl3 is 133.33 g/mol.

H2 is 2.02 g/mol.

4. Determine the molar mass of calcium sulfate [CaSO4].

136.15 g/mol.

5. Determine the molar mass of aluminum nitrate [Al(NO3)3].

213.01 g/mol (each of the three nitrate ions is 62.01 g/mol)

6. Determine the molar mass of aluminum sulfide—what is the correct chemical formula?

Al2S3 is 150.17 g/mol.

7. Determine the molar mass of sodium sulfate [Na2SO4].

142.07 g/mol.

8. Determine the molar mass of calcium acetate.

Ca(CH3COO)2 is 158.18 g/mol.

9. A student obtains 1.8066 x 1024 carbon dioxide molecules. Calculate the mass of this sample. 24  1mol   44.01g  1.8066 x 10 CO2 molecules     132.1 g CO2  6.022x1023 molecules   1mol 

10. A student obtains 64.00 grams of oxygen gas (O2). How many moles of O2 are present? How many oxygen molecules are present?

 1molO2  64.00 g O2   = 2.000 moles O2  32.00g / mol 

 6.022x1023 molecules    24 2.000 moles O2   1.204 x 10 O2 molecules  1mol 

11. A student obtains 50.00 grams of nitrogen gas (N2). How many moles of N2 are present? How many nitrogen molecules are present?

 1molN 2  50.00 g N2   = 1.784 moles N2  28.02g / mol 

 6.022x1023 molecules    24 1.784 moles N2   1.075 x 10 N2 molecules  1mol 

12. A student obtains 12.00 grams of hydrogen gas (H2). How many moles of H2 are present? How many hydrogen molecules are present? How many hydrogen atoms are present? How many protons are present? How many electrons are present?

 1molH 2  12.00 g H2   = 5.941 moles H2  2.02g / mol 

 6.022x1023 molecules    24 5.941 moles H2   3.577 x 10 H2 molecules  1mol 

 2Hatoms  24   24 3.577 x 10 H2 molecules   7.155 x 10 H atoms 1H 2 molecule 

1proton  7.155 x 1024 H atoms   = 7.155 x 1024 protons 1Hatom  1electron  7.155 x 1024 H atoms   = 7.155 x 1024 electrons  1Hatom 

13. A student obtains 100.00 grams of sulfur hexafluoride gas (SF6). How many moles of SF6 are present? How many SF6 molecules are present? How many sulfur atoms are present? How many fluorine atoms are present? How many electrons are present?

 1molSF6  100.00 g SF6   = 0.6846 moles SF6 146.07g / mol 

 6.022x1023 molecules    23 0.6846 moles SF6   4.123 x 10 SF6 molecules  1mol 

 1Satom  23   23 4.123 x 10 SF6 molecules   4.123 x 10 S atoms 1SF6molecule 

 6Fatoms  23   24 4.123 x 10 SF6 molecules   2.474 x 10 F atoms 1SF6molecule 

 70electrons  23   25 4.123 x 10 SF6 molecules   2.886 x 10 electrons 1SF6molecule 

14. A student obtains 50.00 grams of calcium hydroxide. Determine the number of hydroxide ions present in the sample. Determine the numbers of calcium ions present.

 1mol  50.00 grams Ca(OH)2   = 0.6755 moles Ca(OH)2  74.02g 

 6.022x1023 units    23 0. 6755 moles Ca(OH)2   4.068 x 10 Ca(OH)2 units (The term units is  1mol 

used here rather than molecule because Ca(OH)2 is an ionic compound, not a molecule).  1Ca 2ion  23   23 2+ 4.068 x 10 Ca(OH)2 units   4.068 x 10 Ca ions 1Ca(OH ) 2 unit 

 2OH ions  23   23 - 4.068 x 10 Ca(OH)2 units   8.136 x 10 OH ions 1Ca(OH ) 2 unit 